Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

1. Engineering 36 Chp4: MomentMathematics Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Moments are VECTORS • As Described Last Lecture a Moment is a measure of “Twisting Power” • A Moment has Both MAGNITUDE & Direction and can be Represented as a Vector, M, with Normal Vector properties

3. Moments are VECTORS • Describe M in terms of a unit vector, û, directed along the LoA for M • Find the θm by Direction CoSines

4. M = r X F • Magnitude of M measures the tendency of a force to cause rotation of a body about an Axis thru the pivot-Pt O

5. MomentDirection M = r X F • The sense of the moment may be determined by the right-hand rule • If the fingers of the RIGHT hand are curled from the direction of r toward the direction of F, then the THUMB points in the direction of the Moment

6. M = r X F • Combining (1) & (2) yields the Definition of the vector CROSS PRODUCT (c.f. MTH3) • Engineering Mechanics uses the Cross Product to Define the Moment Vector • û is a unit vector directed by the Rt-Hand Rule • θ is the Angle Between the LoA’s for r & F

7. M = r X F → θ by Tail-toTail • When Finding Moment Magnitudes using: • The Angle θ MUST be determined by placing Vectors r & F in the TAIL-to-TAIL Orientation • See Diagram at Right

8. Recall Vector ADDITION Behaved As Algebraic Addition BOTH Commutative and Associative. The Vector PRODUCT Math-Properties do NOT Match Algebra - Vector Products: Are NOT Commutative Are NOT Associative ARE Distributive Cross Product Math Properties

9. Vector Products Of Cartesian Unit Vectors Vector Prod: Rectangular Comps • Vector Product In Terms Of Rectangular Coordinates

10. rxF in 3D Deteriminant Notation • Consider 3D versions of r & F • Taking the Cross Product Yields M • Determinant Notation provides a convenient Tool For the Calculation • Don’t Forget the MINUS sign in the Middle (j)Term • See also TextBook pg123

11. The Moment About a Point OOf The Resultant Of SeveralConcurrent Forces Is Equal To The Sum Of The MomentsOf The Various Forces About The Same Point O Stated Mathematically Varignon’s Theorem • Varignon’s Theorem Makes It Possible To Replace The Direct Determination Of The Moment of a Force F By The Moments of Its Components (which are concurrent)

12. rxF in 3D  Vector Properties • Cartesian CoOrds for a 3D M vector • The Magnitude of a 3D M vector • DirectionCoSines • Unit Vector

13. rxF in 2D r & F in XY Plane • If r & F Lie in the XY Plane, then rz = Fz = 0. Thus the rxF Determinant • So in this case M is confined to the Z-Direction:

14. Consider the CrowBar Below Direction for r in rF • We Want to find the Torque (Moment) About pt-B due to Pull, P, applied at pt-A using rP • We have Two Choices for r: • r points A→B • r points B→A • Which is Correct?

15. We can find the Direction for r by considering the SIGN of the Moment Direction for r in rF • In this case it’s obvious (to me, anyway) that P will cause CLOCKwise Rotation about Pt-B • In the x-y Plane ClockWise Rotation is defined as NEGATIVE • Test rPand rP y x

16. Put r and r into Component form Equal but Opposite Direction for r in rF • Then the two r’s y • Now let x

17. then the rxP calculations noting Direction for r in rF  y  • Thus rB→A is the CORRECT position vector x

18. To Calc the Moment about pt-B use: Direction for r in rF • The position Vector points FROM the PIVOT-point TO the Force APPLICATION-point on the Force LoA • Summarize this as FROM the PIVOTTOthe FORCE

19. Unit Vector Notation: u≡λ • Our Text uses u to denote the unit vector • While u is quite popular as the unit vector notation, other symbols are often used (kind of like θ & φ for angles) • On Occasion I will use λ to represent the unit vector • This is usually apparent from the problem or situation context

20. A Rectangular Plate Is Supported By The Brackets At A and B and By A Wire CD. Knowing That The Tension In The Wire is 200 N, Determine The Moment About A Of The Force Exerted By The Wire At connection-point C. Solution Plan The Moment MA Of The Force F Exerted By The Wire Is Obtained By Evaluating The Vector Product Example: 3D Moment

21. Resolve Both F and rAC into Cartesian Components Take Cross-Product Using Determinant Example 3D Moment - Solution Which Moment will Most Likely Cause DEFORMATION?

22. Moment MO Of A Force F ,Applied at The Point A, About a Point O, Recall Moment About an Axis (§4.5) • Scalar Moment MOL About AnAXIS OL Is The Projection OfThe Moment Vector MO OntoThe OL Axis using the Dot Product MOL it the tendency of the applied force to cause a rotation about the AXIS OL

23. Moments of F About The CoOrd Origin Moment About an Axis – cont. • Moment Of A Force AboutAn Arbitrary Axis BL • Similar Analysis for CL, Starting With MC, Shows That MCL = MBL; i.e., the Result is Independent of the Location of the Point ON the Line

24. Moment About an Axis – cont. • Since the moment, ML,about an arbitrary axisis INDEPENDENT ofposition vector, r, thatruns from ANY Pointon the axis to ANY pointon the LoA of the force we can choose the MOST CONVENIENT Points on the Axis and the Force LoA to determine ML

25. MOL Physical Significance • MOL Measures the Tendency of an Applied Force to Impart to a Rigid Body Rotation about a fixed Axis OL • i.e., How Much will the Applied Force Cause The body to Rotate about an AXLE • MOL can be Considered asthe Componentof M directed along “axis” OL

26. A Cube With Side Length ais Acted On By a Force Pas Shown Determine The Moment Of P: About PtA About The Edge (Axis) AB About The Diagonal (Axis) AGof The Cube For Lines AG and FC Determine The Perpendicular Distance Between them Example: MOL

27. Moment of P about A Example MOL - Solution Moment of P about AB

28. Example MOL - Solution Alternative Moment of P about A

29. Moment of P About Diagonal AG Example MOL - Solution

30. Perpendicular distance between AG and FC Notice That Plane OFC Appears To Be  to Line AG, And FC Resides In this Plane Since P Has Line-of-Action FC We Can Test Perpendicularity with Dot Product Example MOL - Solution  • Then the Moment (or twist) Caused by P About AG = Pd; Thus

31. Do Find MOL we used the Qtyû•(r x F). Formalize thisOperation as the Mixed TripleProduct for vectors S, P, & Q Mixed Triple Product • Associativity and Communtivity for the Mixed Triple Product Of Three Vectors

32. Let V = PxQ, Then Evaluate the Mixed Triple Prod • And • Thus • Determinant Notation Yet Again

33. Mixed Triple Product vs MOL • The Mixed Triple Product can be used to find the Magnitude of the Moment about an Axis.

34. WhiteBoard Work Let’s WorkThis NiceProblem • Determine MA as caused by application ofthe 120 N force

37. TWISTING Power of a Force MOMENT of the Force Quantify Using VECTOR PRODUCT or CROSS PRODUCT Vector/Cross Product • Vector Product Of Two VectorsP And Q Is Defined As TheVector V Which Satisfies: • Line of Action of V Is Perpendicular To Plane Containing P and Q. • Rt Hand Rule Determines Direction for V • |V| =|P|•|Q|•sin